Let v be an eigenvector. Then Av=cv, where c is the corresponding eigenvalue. A^k v=c^k v=0. Then c must be 0. Therefore, there is only one eigenvalue. Now J=P^-1 A P, where J is the Jordan normal form of the matrix and P is a non singular matrix. You can easily check that J^n=P^-1 A^n P. Now, the Jordan normal form is either 0, in which case, Ju=P^-1 A P u=0, for any vector u, which implies that A is 0; or the Jordan normal form is a matrix with all entries 0, except above the diagonal where they are 1's (some authors prefer to put the 1's below the diagonal). Now you can also check that after n multiplications of any vector u with J^n, it will be 0, which implies that A^n is 0.